Answer :
Answer:
The work done on the cord is [tex]2.845\times10^{4}\ J[/tex].
Explanation:
Given that,
The function
[tex]F(z)=k_{1}z+k_{2}[/tex]
Spring constant
[tex]K_{1}=204 N/m[/tex]
[tex]K_{2}=0.233\ N/m[/tex]
Stretch length b = 16.7 m
We need to calculate the work done
Using formula of work done
[tex]W=\int_{a}^{b}{F(z)}dz[/tex]
Put the value into the formula
[tex]W=\int_{a}^{b}{k_{1}z+k_{2}}dz[/tex]
[tex]W=\int_{a}^{b}{k_{1}zdz+k_{2}dz}[/tex]
On integrating
[tex]W=({\dfrac{204\times z^2}{2}})_{0}^{16.7}+({0.233\times z})_{0}^{16.7}[/tex]
[tex]W={\dfrac{204\times16.7^2}{2}+0.233\times16.7}[/tex]
[tex]W=2.845\times10^{4}\ J[/tex]
Hence, The work done on the cord is [tex]2.845\times10^{4}\ J[/tex].
The work that must be done on the code to stretch it is 28,450.7 J.
Work done in stretching the cord
The work done in stretching the cord is obtained from the integral of the non elastic function as follows;
W = ∫F(z) + Kadz
W = ∫K₁z + Kadz
W = ¹/₂K₁z² + Kaz
Where;
- K₁ = 204 N/m
- Ka = 0.233 N/m
- Z = 16.7
W = ¹/₂ x 204 x (16.7)² + 0.233 X 16.7
W = 28,450.7 J
Thus, the work that must be done on the code to stretch it is 28,450.7 J.
Learn more about work done here: https://brainly.com/question/8119756